Question

What is the theoretical yield for the reaction between 6 g of aluminum and an excess of bromine, resulting in 50.3 g of aluminum bromide?

a. 40.5 g
b. 60.8 g
c. 75.2 g
d. 100.6 g

Answer 1

The theoretical yield for the reaction between 6 g of aluminum and an excess of bromine, resulting in 50.3 g of aluminum bromide, is 57.64 g.

Step-by-step explanation:

The theoretical yield for the reaction between 6 g of aluminum and an excess of bromine, resulting in 50.3 g of aluminum bromide, can be calculated using stoichiometry. The reaction's balanced equation is 2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s).

From the equation, we can see that 2 moles of Al produce 2 moles of AlBr3. Therefore, if we have 6 g of Al, which is equivalent to 0.216 moles, we will produce 0.216 moles of AlBr3. The molar mass of AlBr3 is 266.69 g/mol, so the theoretical yield is calculated as follows:

Theoretical Yield = 0.216 moles x 266.69 g/mol = 57.64 g

Therefore, the correct answer is a. 57.64 g.